Calculating Velocity of Sphere on Inclined Plane

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SUMMARY

The discussion focuses on calculating the velocity of a sphere as it rolls up a 25-degree incline after traveling at a constant speed of 10 m/s on a horizontal surface. The sphere has a mass of 25 kg and a radius of 0.2 m, with a moment of inertia defined as I=2mr²/5. The participant initially calculated the total kinetic energy as 3015 J by combining translational kinetic energy, rotational kinetic energy, and gravitational potential energy (mgh). However, confusion arose regarding the increase in velocity while ascending the incline, leading to a request for clarification on the energy calculations.

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  • Understanding of translational and rotational kinetic energy
  • Familiarity with the moment of inertia formula for solid spheres
  • Knowledge of gravitational potential energy calculations
  • Basic principles of energy conservation in physics
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Homework Statement


A large sphere rolls wothout slipping across a horizontal surface. The sphere has a constant translational speed of 10m/s,mass of 25 kg, radius of .2m. The moment of inertia of the sphere about its center of mass is I=2mr2/5. The sphere approaches a 25 degree incline of height 3m as shown above and rolls up the incline without slipping. Calculate the spheres velocity just as it leaves the top of the incline.

Homework Equations





The Attempt at a Solution


I found total kinetic energy by adding translational kin enery, rotational kin energy and (mgh)... i got 3015J and then set that number equal to the equation for kin energy (1/2mv2) and found my new v and got 15.53 m/s but that can't be correct bc if its initial vel as it was going toward the ramp was 10 m/s then why would it speed up if its going up a ramp? idk what i did wrong pleaseee help.!
 
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moondawg said:
I found total kinetic energy by adding translational kin enery, rotational kin energy and (mgh)...
Why did you add mgh? Explain exactly what you did.
 

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